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I want to show, that $sinh(R)=R$.

As $ sinh = \frac{e^x-e^{-x}}{2}$ I could say, that $e^x$ is always an element of $R$, if $x∈ R$, so as $e^{-x}=\frac{1}{e^x}∈ R$. Substracting a real number from another real number we get a real number again and dividing a real number with $2$, we get a real number again.

Is there any smarter way, which is mathematically better to show our assumption? Should I also prove that $e^x∈ R$ if $x∈ R$ ?

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I think you are confused. $\sinh(\mathbb{R})=\mathbb{R}$ means that $\sinh$ is surjective, i.e. that every real number is of the form $\sinh(x)$ for some $x\in\mathbb{R}$. The fact that $\sinh(x)$ is real for $x\in\mathbb{R}$ is trivial: it's because it's a composition of real functions.

In order to show that $\sinh$ is surjective all you need to do is to prove these:

$$\lim_{x\to -\infty} \frac{e^{x}-e^{-x}}{2} = -\infty$$ $$\lim_{x\to \infty} \frac{e^{x}-e^{-x}}{2} = \infty$$

Since $\sinh$ is continous then it has the intermediate value property. Therefore $\sinh$ is surjective.

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    thank you @freakish, you are right, I was confused. is it also possible to prove that sinh is continous? – 2017-01-27
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    @ÁrminBéda $e^x$ is continous, $-x$ is continous, $\frac{1}{2}x$ is continous. Composition of continous functions is continous. Adding continous functions is continous. – 2017-01-27